【luogu P2194 HXY燒情侶】 題解
阿新 • • 發佈:2018-07-27
ns2 原理 str struct code add mem problem 題目
題目鏈接:https://www.luogu.org/problemnew/show/P2194
第一問:縮點並且統計其強連通分量裏的最小耗費。把所有強連通分量的最小耗費加起來。
第二問:統計在每個強連通分量裏與最小耗費相同的點數。乘法原理統計所有強連通分量答案。
#include <stack> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 300000 + 10; const int inf = 0x7fffffff; const int mod = 1e9 + 7; struct edge{ int from, to, next; }e[maxn<<2]; int head[maxn], cnt; int n, m, ans1, ans2 = 1, dfn[maxn], low[maxn], tim, color[maxn], num, val[maxn], minpay[maxn], tot[maxn]; bool vis[maxn]; stack<int> s; void add(int u, int v) { e[++cnt].from = u; e[cnt].next = head[u]; e[cnt].to = v; head[u] = cnt; } void tarjan(int x) { dfn[x] = low[x] = ++tim; vis[x] = 1; s.push(x); for(int i = head[x]; i != -1; i = e[i].next) { int v = e[i].to; if(!dfn[v]) { tarjan(v); low[x] = min(low[x], low[v]); } else if(vis[v]) { low[x] = min(low[x], low[v]); } } if(dfn[x] == low[x]) { color[x] = ++num; vis[x] = 0; minpay[num] = min(minpay[num], val[x]); while(s.top() != x) { color[s.top()] = num; vis[s.top()] = 0; minpay[num] = min(minpay[num], val[s.top()]); s.pop(); } s.pop(); } } int main() { memset(head, -1, sizeof(head)); scanf("%d",&n); for(int i = 1; i <= n; i++) minpay[i] = inf; for(int i = 1; i <= n; i++) scanf("%d",&val[i]); scanf("%d",&m); for(int i = 1; i <= m; i++) { int u, v; scanf("%d%d",&u,&v); add(u,v); } for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i); //for(int i = 1; i <= n; i++) cout<<minpay[i]; for(int i = 1; i <= num; i++) ans1 += minpay[i]; cout<<ans1<<" "; for(int i = 1; i <= n; i++) { if(val[i] == minpay[color[i]]) tot[color[i]]++; } for(int i = 1; i <= num; i++) ans2 = (ans2*tot[i])%mod; cout<<ans2<<endl; return 0; }
【luogu P2194 HXY燒情侶】 題解